Giải thích các bước giải:

a.Do $\Delta ABC$ cân tại A, gọi H là trung điểm BC$\rightarrow AH\perp BC$

$\rightarrow AB=AC=m, \hat{B}=\hat{C}=90-\dfrac{\alpha}{2}$ 

$\rightarrow sin\dfrac{\alpha}{2}=\dfrac{BH}{AB}=\dfrac{BH}{m}$

$\rightarrow BH=m.sin\dfrac{\alpha}{2}$

$\rightarrow AH=\sqrt{AB^2-BH^2}=m.cos\dfrac{\alpha}{2}$

$\rightarrow BC=2BH=2m.sin\dfrac{\alpha}{2}$

$\rightarrow DH=\dfrac{BC}{6}=\dfrac{m.sin\dfrac{\alpha}{2}}{6}$

$\rightarrow AD^2=AH^2+DH^2=(\dfrac{m.sin\dfrac{\alpha}{2}}{6})^2+(m.cos\dfrac{\alpha}{2})^2$

$\rightarrow AD=m\sqrt{(\dfrac{sin\dfrac{\alpha}{2}}{6})^2+(cos\dfrac{\alpha}{2})^2}$

b.Áp dụng định lý sinh ta có:

$R_{ABD}=\dfrac{AD}{2sin\hat{B}}=\dfrac{AC}{2sin\hat{C}}=R_{ACD}$

Để $R_{ABD}=\dfrac{1}{2}R_{ABC}$

$\rightarrow \dfrac{AD}{2sin\hat{B}}=\dfrac{1}{2}.\dfrac{AC}{2sin\hat{B}}$

$\rightarrow AD=\dfrac{1}{2}.AC=\dfrac{1}{2}m$

$\rightarrow m\sqrt{(\dfrac{sin\dfrac{\alpha}{2}}{6})^2+(cos\dfrac{\alpha}{2})^2}=\dfrac{1}{2}m$

$\rightarrow (\dfrac{sin\dfrac{\alpha}{2}}{6})^2+(cos\dfrac{\alpha}{2})^2=\dfrac{1}{4}$

$\rightarrow (\dfrac{sin\dfrac{\alpha}{2}}{6})^2+1-(sin\dfrac{\alpha}{2})^2=\dfrac{1}{4}$

$\rightarrow sin\dfrac{\alpha}{2}=\sqrt{\dfrac{27}{35}}$

$\rightarrow \alpha=2.arcsin\sqrt{\dfrac{27}{35}}$